Advanced Data Structures

Transcription

1 Simon Gog - Simon Gog: KIT The Research University in the Helmholtz Association

2 Predecessor data structures We want to support the following operations on a set of integers from the domain U = [u]. insert(x) Add x to S. I.e. S = S {x}. delete(x) Delete x from S. I.e. S = S \ {x}. member(x) = {x x S} predecessor(x) = max{y y x y S} successor(x) = min{y y x y S} where x is an integer in U and S the set of integers of size n stored in the data structure. min{s} = successor() max{s} = predecessor(u 1) Solution know from Algo I : Balanced search trees. E.g. red-black trees. In all comparison based approaches at least one operation takes Ω(log n) time. Why? 1 Simon Gog:

3 Predecessor data structures Ω(log n) bound can be beaten in the word RAM model: Memory is organized in words of b O(log u) bits A word can be accessed in constant time We can address all data using one word Standard arithmetic operations take constant time on words (i.e. addition, subtraction, division, shifts...) We first concentrate on the static case: The set S is fixed. I.e. no insert and delete operations. 2 Simon Gog:

4 x-fast trie (Willard, 1982) Conceptional: Complete binary tree of height w = log u Operations member/successor/prodecessor can be answered in O(w) time by traversing the tree 3 Simon Gog:

5 x-fast trie (Willard, 1982) Conceptional: Complete binary tree of height w = log u Operations member/successor/prodecessor can be answered in O(w) time by traversing the tree 3 Simon Gog:

6 x-fast trie (Willard, 1982) Conceptional: Complete binary tree of height w = log u Operations member/successor/prodecessor can be answered in O(w) time by traversing the tree 3 Simon Gog:

7 x-fast trie (Willard, 1982) Conceptional: Complete binary tree of height w = log u Operations member/successor/prodecessor can be answered in O(w) time by traversing the tree 3 Simon Gog:

8 x-fast trie From O(log u) to O(log log u) Simon Gog:

9 x-fast trie From O(log u) to O(log log u) Generate a perfect hash table h i for each level i (keys are the present nodes) Note: Node numbers (represented in binary) are prefixes of keys in S 5 Simon Gog:

10 x-fast trie From O(log u) to O(log log u)... h h 1 1 h h h Generate a perfect hash table h i for each level i (keys are the present nodes) Note: Node numbers (represented in binary) are prefixes of keys in S 5 Simon Gog:

11 x-fast trie Query time from O(log u) to O(log log u)... Member queries can be answered in constant time by lookup in the leaf level using prefixes of the searched key. There are w prefixes, i.e. binary search takes O(log w) or O(log log u) time Space: w O(n) words, i.e. O(n log u) bits Predecessor/Successor queries For each node store a pointer to the maximal/minimal leaf in its subtree Use a double linked list to represent leaf nodes. Solving predecessor: Search for a node v which represents the longest prefix of x with any key in S. Two cases: Minimum in subtree of v is larger x: Return element to the left of the leaf. Maximum in the subtree of v is smaller than x. Return maximum. 6 Simon Gog:

12 y-fast trie (Willard, 1982) Space from O(n log u) to O(n) words... Split S into n w blocks of O(log u) elements B, B 1,..., B n w 1. max{b i } < min{b i+1 } for i < n w 1 Let r i = max{b i } be a representative of block B i. Build x-fast trie over representatives. B B 1... Total space: w n O(w) + O(n) = O(n) words 7 Simon Gog:

13 y-fast trie Space from O(n log u) to O(n) words... Use sorted array to represent B i A member query is answered as follows Search for successor of x in x-trie of r i s Let B k be the block of the successor of x Search in O(log w) = O(log log u) time for x in B k How does predecessor/successor work? 8 Simon Gog:

14 y-fast trie Changes to make structure dynamic use cuckoo hashing for x-fast trie use balanced search trees of size between 1 2 w and 2w for B is representative is not the maximum, but any element separating two consecutive groups Summary: Operation static y-fast trie dynamic y-fast trie pred(x)/succ(x) O(log log u) w.c. O(log log u) insert(x)/delete(x) O(log log u) exp. & am. construction O(n) exp. 9 Simon Gog:

15 Van Emde Boas Trees Conceptual bitvector B of length u with B[i] = 1 for all i S Split B into u/ u blocks (blue blocks) B, B 1,... Set bit in R[i] if there is at least one bit set in B i Also store the minimum/maximum of S Summary R min = 1 max = 14 B= Here: u = Simon Gog:

16 Van Emde Boas Trees Define van Emde Boas tree (veb tree) recursively I.e. use veb to represent B i s and R veb(u) denotes the veb tree on a universe of size u Base case: u = 2. Only one node and variables min, max. Technicalities u = 2 (log u)/2 u = 2 (log u)/2 high(x) = x (block that contains x) u low(x) = x mod u (relative position of x in B high(x) ) 11 Simon Gog:

17 Van Emde Boas Trees Predecessor(x, B) (first attempt) Let y = height(x) and z = Predecessor(low(x), B y ) If z = return z + y u Let b = Predecessor(high(x), R) If b = return max(b b ) + b u Return Summary R B= 1 B 1 B 1 1 B B Simon Gog:

18 Van Emde Boas Trees Predecessor(x, B) (first attempt) Let y = height(x) and z = Predecessor(low(x), B y ) If z = return z + y u Let b = Predecessor(high(x), R) If b = return max(b b ) + b u Return Problem Recurrence for time complexity: T (u) = 2T ( u) + Θ(1) Substitute u by 2 k and define S(k) = T (2 k ) S(k) = 2S( k 2 ) + Θ(1) (solved by Master Theorem) We get T (u) = O(log u log log u). Next: improve time to O(log log u) 12 Simon Gog:

19 Van Emde Boas Trees Predecessor(x, B) (first attempt) Let y = height(x) and z = Predecessor(low(x), B y ) If z = return z + y u Let b = Predecessor(high(x), R) If b = return max(b b ) + b u Return Problem Recurrence for time complexity: T (u) = 2T ( u) + Θ(1) Substitute u by 2 k and define S(k) = T (2 k ) S(k) = 2S( k 2 ) + Θ(1) (solved by Master Theorem) We get T (u) = O(log u log log u). Next: improve time to O(log log u) 12 Simon Gog:

20 Van Emde Boas Trees Predecessor(x, B) (first attempt) Let y = height(x) and z = Predecessor(low(x), B y ) If z = return z + y u Let b = Predecessor(high(x), R) If b = return max(b b ) + b u Return Problem Recurrence for time complexity: T (u) = 2T ( u) + Θ(1) Substitute u by 2 k and define S(k) = T (2 k ) S(k) = 2S( k 2 ) + Θ(1) (solved by Master Theorem) We get T (u) = O(log u log log u). Next: improve time to O(log log u) 12 Simon Gog:

21 Van Emde Boas Trees Predecessor(x, B) (second attempt) If x > max return max Let y = high(x), if min(b y ) < x return Predecessor(low(x), B y ) Let b = Predecessor(high(x), R) If b = return max(b b ) + b u Return Recurrence for time complexity: T (u) = T ( u) + O(1) Solution (Master Theorem or drawing recursion tree): T (u) = Θ(log log u) 13 Simon Gog:

22 Van Emde Boas Tree Space complexity Note Recurrence: S(u) = ( u + 1)S( u) + Θ(1) Solution: S(u) O(u) Space complexity of x-fast and y-fast tries are better for small sets Van Emde Boas Tree does not rely on hashing 14 Simon Gog: